A police car is traveling at a velocity of 20.0 m/s due north, when a car zooms by at a constant velocity of 44.0 m/s due north. After a reaction time 0.700 s the policeman begins to pursue the speeder with an acceleration of 6.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?
44m/s * 0.7s = 30.8 m. Lead.
d2 = d1 + 14 m.
Vo*t + 0.5a*t^2 = 44t + 14
20t + 3*t^2 = 44t + 14
3t^2 - 24t - 14 = 0
Use Quadratic formula.
t = 8.55 s.
Correction:
44m/s * 0.7s = 30.8 m. Lead.
d2 = d1 + 30.8
Vo*t + 0.5a*t^2 = 44t + 30.8
20t + 3t^2 = 44t + 30.8
3t^2 - 24t - 30.8 = 0
Use Quad. Formula.
t = 9.125 s.
To find the time it takes for the police car to catch up with the speeder, we need to analyze the motion of both vehicles.
Let's first consider the speeder's motion. The initial velocity of the speeder (v1) is 44.0 m/s due north, and we can assume it remains constant. Therefore, the equation for the speeder's displacement (s1) at any given time (t) is:
s1 = v1 * t
Next, let's analyze the police car's motion. The initial velocity of the police car (v0) is 20.0 m/s due north. However, we need to consider the reaction time (tr) of 0.700 s before the police car starts accelerating. During this time, the police car continues to move at a constant velocity. Therefore, we can find the displacement of the police car during reaction time (sr):
sr = v0 * tr
After the reaction time, the police car starts accelerating with an acceleration (a) of 6.00 m/s^2. Using the equations of motion, we can find the displacement of the police car (s0) in terms of time (t) after the reaction time:
s0 = v0 * t + (1/2) * a * t^2
Now, to find the time it takes for the police car to catch up with the speeder, we need to find the time at which both vehicles have traveled the same displacement. In other words, when s0 = s1.
v0 * t + (1/2) * a * t^2 = v1 * t + sr
Substituting the given values:
20.0 * t + (1/2) * 6.00 * t^2 = 44.0 * t + (20.0 * 0.700)
Now we have a quadratic equation that we can solve to find the value of t.
Since this is a quadratic equation, we can rearrange it to the standard form:
(1/2) * 6.00 * t^2 - 24.0 * t + (20.0 * 0.700) = 0
Now we can solve this equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-(-24.0) ± sqrt((-24.0)^2 - 4 * (1/2) * (20.0 * 0.700))) / (2 * (1/2) * 6.00)
Simplifying further:
t = (24.0 ± sqrt(576.0 - 56.0)) / 3.00
t = (24.0 ± sqrt(520.0)) / 3.00
Finally, calculating the values:
t ≈ (24.0 + sqrt(520.0)) / 3.00 ≈ 6.965 s
t ≈ (24.0 - sqrt(520.0)) / 3.00 ≈ -1.632 s (discard negative time)
Therefore, it takes approximately 6.965 seconds (including the reaction time) for the police car to catch up with the speeder.